Proof: Let $x^{2}=u^{3}, y^{2}=v^{3}, z^{2}=w^{3}$, then equation (36) can be rewritten as
$$\frac{v w}{v w+2 u^{2}}+\frac{w u}{w u+2 v^{2}}+\frac{u v}{u v+2 w^{2}} \geqslant 1,$$

where $u, v, w$ are positive real numbers, and satisfy $u v w=4$.
Since
$$\begin{aligned} & \frac{v w}{v w+2 u^{2}}+\frac{w u}{w u+2 v^{2}}+\frac{u v}{u v+2 w^{2}}-1 \\ = & \frac{u v w(u+v+w)\left[(v-w)^{2}+(w-u)^{2}+(u-v)^{2}\right]}{\left(2 u^{2}+v w\right)\left(2 v^{2}+w u\right)\left(2 w^{2}+u v\right)} \geqslant 0 \end{aligned}$$

Therefore, equation (37) holds, which means equation (36) holds.